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calculate the standard emf for each of the following reactions. H2(g)+F2(g)→2H+(aq)+2F−(aq). Cu2+(aq)+Ca(s)→Cu(s)+Ca2+(aq).
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In tables, all reaction are reduction reactions For H : 2 H+ + 2 e- -> H2 E = 0 V (definition) In first reaction H oxidized, and E = 0 V Then find E for reaction F2 + 2e -> 2 F- calculate sum together In next reaction Copper is reduced and Ca oxidized Find E for reactions Cu2+ + 2e- -> Cu Ca2+ + 2e- -> Ca Because lower reaction happens from right to left, you have to change sign for E.
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